Last name:
name:
1
Quiz 9 (Notes, books, and calculators are not authorized) Show all your work in the blank space
you are given on the exam sheet. Answers with no justification will not be graded.
Question 1: (a) Find the derivative of f (z) = 5z 3 +3iz−3+i and specify where f is holomorphic
f 0 (z) = 15z 2 + 3i.
f is holomorphic in C.
(b) Find the derivative of F (z) =
2z+i
z+2i
and specify where F is holomorphic.
2(z + 2i) − (2z + i)
3i
=
.
2
(z + 2i)
(z + 2i)2
F 0 (z) =
F is holomorphic in C \ {− 21 i}.
Question 2: Let f (z) = z 3 . Compute integral
arc of radius 1 starting at 1 and ending at −1.
R
C
f (z)dz where C is the upper semi-circular
solution 1: Owing to a corollary of Cauchy’s Theorem the integral is independent of the path that
is chosen to connect 1 to −1. We may choose the line γ(t) = 1 − 2t, t ∈ [0, 1], then
Z
Z
f (z)dz =
C
1
−2t3 dt = 0
0
solution 2: Another possibility consists of realizing that f (z) = g 0 (z) where g(z) = 41 z 4 , then
Z
Z
f (z)dz =
g 0 (z)dz = g(−1) − g(1) = 0
C
C
solution 3: Compute the integral using the definition of the semi-circular arc: γ(t) = eiπt , with
t ∈ [0, 1].
Z
Z
f (z)dz =
C
0
1
e3iπt iπeiπt dt ==
1
4
Z
0
1
4iπe4iπt dt =
1 4iπ
(e − 1) = 0.
4
2
Quiz 9, November 29, 2012
Question 3: Compute
−5 + 2i.
R
dz
C z−2
where C is the perimeter of the square of width 3 centered at
1
is holomorphic in the open set C \ {2} and the square of width 3 centered
The function h(z) = z−2
at −5 + 2i is in C \ {2}. Then the integral is zero owing to Cauchy’s Theorem.
R √
√
√ θ
Question 4: Let us define z = rei 2 for θ ∈ [−π, π], where z = reiθ . Compute C zdz
where C is the straight line connecting 1 − i to 1 + i.
√
Solution 1: The definition of z implies that this function is holomorphic in the complex plane
minus the axis (−∞, 0). Then
Z
Z
√
√
zdz =
zdz,
C
√
L
where L is the portion of circular arc of radius 2 centered√at 0 and joining 1 − i to 1 + i. A
π
π
possible parameterization of L is defined as follows: γ(t) = 2ei(t 2 − 4 ) , with t ∈ [0, 1]. Observe
π
π
that t 2 − 4 ∈ [−π, π] when t spans [0, 1]. Then
Z
C
√
Z
1
π
π
π
2 3 3π
2 i ei(t 2 − 4 ) dt = 2 4 i
zdz =
2 e
2
3
4
0
3π
2 3 −i 3π i 3π
2 3 i 3π
= 2 4 e 8 (e 4 − 1) = 2 4 (e 8 − e−i 8 ).
3
3
1
4
π
i(t π
4−8)
1
2
Z
1
ei(t
3π
3π
4 − 8 )
0
In conclusion
Z
C
√
4 3
3π
zdz = i 2 4 sin( )
3
8
3
√
Solution 2: Observe that z = 23 dzdz2 . Hence
Z
√
3
3
3π
2
2 3 3π
zdz = ((1 + i) 2 − (1 − i) 2 ) = 2 4 (ei 8 − e−i 8 ).
3
3
C
dt